1) A hydrogen atom and a neutron fuse to form deuterium. If the energy released

Question

1) A hydrogen atom and a neutron fuse to form deuterium. If the energy released is 3.6 x 10-13 J, what is the mass of the deuterium? Take the H atom nucleus as a single proton.

Explanation / Answer

We assume that the following reaction holds: (proton) + (neutron) ------> deuterium + energy. The energy comes from mass decay according to E = mc^2. Or, m = E/c^2. We have: E = 3.6*10^-13 J c = 3.0 * 10^8 m/s ---> m = 3.6*10^-13/(3.0 * 10^8)^2 = 4.00 * 10^-30 kg. This is the mass lost during fission. Now, since: m (proton) = 1.6726 * 10^-27 kg m (neutron) = 1.6749 * 10^-27 kg. We have: mass (deuterium) = 1.6726*10^-27 + 1.6749*10^-27 - 4.00 * 10^-30 = 3.3435*10^-27 kg = 2.0135 amu.

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